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bezimeni [28]
3 years ago
9

Which one of the atoms shown would be most likely to form a cation with a charge of 1?

Chemistry
2 answers:
olga55 [171]3 years ago
6 0
Ions are stable atoms of elements formed when original atoms loose or gain electrons to form stable configuration. Anions are negatively charged ions formed when atoms gain electrons while cations are the positively charged ions formed when atoms loose electrons. In this case a charge of +1 means an atom lost one electron to form a stable configuration. Therefore, the correct answer is A
Dafna1 [17]3 years ago
6 0
If I am referring to correct picture then  
A. Has Four atoms in outer ring 
B. Has Five atoms in outer ring 
C. Has Six atoms in outer ring 
D. Has Seven atoms in outer ring 
E. Has Eight atoms in outer ring 
Answer is A 
Now we know that an atom wants to complete its outer shell while keeping electrons in pairs of two now in A there are four electrons which which can be ejected while in B will want to accept 3 electrons to complete its shell as ejecting five will take lot of energy similar case will be for C,D and E which would want to accept 2,1,0 electrons respectively
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Nitella [24]
Why did you post this again? Because you're lazy and wanted people to not see my post? Once again, ONE QUESTION AT A TIME. We're here to help you with a tough question or work you through it, not do all your homework.
8 0
3 years ago
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The reaction for the formation of gaseous hydrogen fluoride (HF) from molecular hydrogen (H2) and fluorine (F2) has an equilibri
mestny [16]

Answer:

[H2]eq = 0.0129 M

[F2]eq = 1.0129 M

[HF]eq = 0.9871 M

Explanation:

  • H2(g) + F2(g) ↔ 2HF(g)

∴ Ke = [HF]² / [H2]*[F2] = 1.15 E2

experiment:

∴ n H2 = 3.00 mol

∴ n F2 = 6.00 mol

∴ V sln = 3.00 L

⇒ [H2]i = 3.00 mol / 3.00 L = 1 M

⇒ [F2]i = 6.00 mol / 3.00 L = 2 M

        [ ]i    change      [ ]eq

H2     1         1 - x         1 - x

F2     2        2 - x         2 - x

HF     -            x              x

⇒ K = (x)² / (1 - x)*(2 - x) = 1.15 E2

⇒ x² / (2 - 3x + x²) = 1.15 E2 = 115

⇒ x² = (2 - 3x + x²)(115)

⇒ x² = 230 - 345x + 115x²

⇒ 0 = 230 - 345x + 114x²

⇒ x = 0.9871

equilibrium:

⇒ [H2] = 1 - x = 1 - 0.9871 = 0.0129 M

⇒ [F2] = 2 - x = 2 - 0.9871 = 1.0129 M

⇒ [HF] = x = 0.9871 M

5 0
3 years ago
A mixture initially contains AA, BB, and CC in the following concentrations: [A][A]A_1 = 0.550 MM , [B][B]B_1 = 1.40 MM , and [C
Alex787 [66]

Answer:

The value of the equilibrium constant KC is 1.244

Explanation:

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.550 M, [B] = 1.40 M, and [C] = 0.600 M. The following reaction occurs and equilibrium is established: A+2B<->C

At equilibrium, [A] = 0.430 M and [C] = 0.720 M. Calculate the value of the equilibrium constant, Kc

Step 1: The balanced equation

A+2B<->C

Step 2: The initial concentrations

[A] = 0.550 M

[B]= 1.40 M

[C] = 0.600 M

Step 3: The concentraions at equilibrium

[A] = 0.550 -X = 0.430 M

[B]= 1.40 -2X M

[C] = 0.600 + X = 0.720 M

X = 0.120 M

[A] = 0.550 - 0.120 = 0.430 M

[B]= 1.40 -2*0.120 =  1.16 M

[C] = 0.600 + 0.120 = 0.720 M

Step 4: Calculate Kc

Kc = [C] / [A][B]²

Kc = 0.720 / (0.430*1.16²)

Kc = 1.244

The value of the equilibrium constant KC is 1.244

5 0
3 years ago
A gold-colored ring has a mass of 17.5 grams and a volume of 0.82 mL. What is the density of this ring?
max2010maxim [7]

Answer:

21 g/mL

Explanation:

To solve this problem, first look at the density equation, which is D=M/V, which D stands for density, M stands for mass, and V stands for volume. When you substitute in the variables, you get D=17.5/.82, which is equivalent to 21.34. However, since we need to pay attention to the sig fig rules for multiplying, we need to have the same amount of sig figs as the value with the least amount of sig figs, which is the number .82. .82 has two sig figs, so you round down. Your answer will be 21 g/mL.

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3 years ago
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